When is instantaneous and average rate of change the same?

Given the function f(x)= (x-2) / (x-5), determine an interval and a point where the ave. R.O.C and the instantaneous R.O.C are equal.

IROC = [ f(x+h) - f(x) ] /h
AROC = f(x2) - f(x1) / x2 - x1

I know that in order to satisfy this, the x's must satisfy both equations when using the AROC and IROC formulas.
I'm not sure where to start, though.
I could do guess and check, but I need help for an algebraic method.

This can be seen from Lagrange's Mean Value Theorem.
in fact there should exist infinitely many such intervals.
Choose any interval [a,b] which satisfies continuity of the function in [a,b] and differentiability in (a,b) then you will always get a c $\displaystyle \epsilon$ (a,b) such that:
f'(c)=[f(b)-f(a)]/(b-a)

eg: a=6, b=7
AROC=(2.5-4)/(1)=-1.5
$\displaystyle f'(x)= (-3)/[(x-5)^2]=-1.5$
then we have $\displaystyle (x-5)^2=2$
so $\displaystyle x=5+sqrt(2)$ which of course belongs to [6,7]